The NCERT solutions for Class 8 Maths enhance topics with frequent, focused, engaging maths challenges and activities that strengthen maths concepts. Each question of exercise 6.1 in NCERT Class 8 Maths Solutions has been carefully solved for the students to understand, keeping the examination point of view in mind.
Class 8 Maths Chapter 6 – Squares and Square Roots Exercise 6.1 Questions and answers help students to understand the difference between squares and square roots, as well as how to find them out. These NCERT Solutions are prepared by subject experts at BYJU’S using a step-by-step approach.
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Exercise 6.2 Solutions 2 Questions
Exercise 6.3 Solutions 10 Questions
Exercise 6.4 Solutions 9 Questions
Access Answers of Maths NCERT Class 8 Chapter 6 – Squares and Square Roots Exercise 6.1 Page Number 96
1. What will be the unit digit of the squares of the following numbers?
i. 81
ii. 272
iii. 799
iv. 3853
v. 1234
vi. 26387
vii. 52698
viii. 99880
ix. 12796
x. 55555
Solution:
The unit digit of the square of a number having ‘a’ at its unit place ends with a×a.
i. The unit digit of the square of a number having digit 1 as the unit’s place is 1.
∴ The unit digit of the square of the number 81 is equal to 1.
ii. The unit digit of the square of a number having the digit 2 as the unit’s place is 4.
∴ The unit digit of the square of the number 272 is equal to 4.
iii. The unit digit of the square of a number having the digit 9 as the unit’s place is 1.
∴ The unit digit of the square of the number 799 is equal to 1.
iv. The unit digit of the square of a number having the digit 3 as the unit’s place is 9.
∴ The unit digit of the square of the number 3853 is equal to 9.
v. The unit digit of the square of a number having the digit 4 as the unit’s place is 6.
∴ The unit digit of the square of the number 1234 is equal to 6.
vi. The unit digit of the square of a number having the digit 7 as the unit’s place is 9.
∴ The unit digit of the square of the number 26387 is equal to 9.
vii. The unit digit of the square of a number having the digit 8 as the unit’s place is 4.
∴ The unit digit of the square of the number 52698 is equal to 4.
viii. The unit digit of the square of a number having the digit 0 as the unit’s place is 01.
∴ The unit digit of the square of the number 99880 is equal to 0.
ix. The unit digit of the square of a number having the digit 6 as the unit’s place is 6.
∴ The unit digit of the square of the number 12796 is equal to 6.
x. The unit digit of the square of a number having the digit 5 as the unit’s place is 5.
∴ The unit digit of the square of the number 55555 is equal to 5.
2. The following numbers are obviously not perfect squares. Give reason.
i. 1057
ii. 23453
iii. 7928
iv. 222222
v. 64000
vi. 89722
vii. 222000
viii. 505050
Solution:
We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.
i. 1057 ⟹ Ends with 7
ii. 23453 ⟹ Ends with 3
iii. 7928 ⟹ Ends with 8
iv. 222222 ⟹ Ends with 2
v. 64000 ⟹ Ends with 0
vi. 89722 ⟹ Ends with 2
vii. 222000 ⟹ Ends with 0
viii. 505050 ⟹ Ends with 0
3. The squares of which of the following would be odd numbers?
i. 431
ii. 2826
iii. 7779
iv. 82004
Solution:
We know that the square of an odd number is odd, and the square of an even number is even.
i. The square of 431 is an odd number.
ii. The square of 2826 is an even number.
iii. The square of 7779 is an odd number.
iv. The square of 82004 is an even number.
4. Observe the following pattern and find the missing numbers. 112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 …….2………1
100000012 = ……………………..
Solution:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1 and the middle digit is 2. And the number of zeros between the left-most digit 1 and the middle digit 2 and the right-most digit 1 and the middle digit 2 is the same as the number of zeros in the given number.
∴ 1000012 = 10000200001
100000012 = 100000020000001
5. Observe the following pattern and supply the missing numbers. 112 = 121
1012 = 10201
101012 = 102030201
10101012 = ………………………
…………2 = 10203040504030201
Solution:
We observe that the square on the number on R.H.S of the equality has an odd number of digits such that the first and last digits both are 1. And the square is symmetric about the middle digit. If the middle digit is 4, the number to be squared is 10101, and its square is 102030201.
So, 10101012 =1020304030201
1010101012 =10203040505030201
6. Using the given pattern, find the missing numbers. 12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + _2 = 212
5 + _ 2 + 302 = 312
6 + 7 + _ 2 = _ 2
Solution:
Given, 12 + 22 + 22 = 32
i.e 12 + 22 + (1×2 )2 = ( 12 + 22 -1 × 2 )2
22 + 32 + 62 =72
∴ 22 + 32 + (2×3 )2 = (22 + 32 -2 × 3)2
32 + 42 + 122 = 132
∴ 32 + 42 + (3×4 )2 = (32 + 42 – 3 × 4)2
42 + 52 + (4×5 )2 = (42 + 52 – 4 × 5)2
∴ 42 + 52 + 202 = 212
52 + 62 + (5×6 )2 = (52+ 62 – 5 × 6)2
∴ 52 + 62 + 302 = 312
62 + 72 + (6×7 )2 = (62 + 72 – 6 × 7)2
∴ 62 + 72 + 422 = 432
7. Without adding, find the sum.
i. 1 + 3 + 5 + 7 + 9
Solution:
Sum of first five odd numbers = (5)2 = 25
ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19
Solution:
Sum of first ten odd numbers = (10)2 = 100
iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Solution:
Sum of first thirteen odd numbers = (12)2 = 144
8. (i) Express 49 as the sum of 7 odd numbers.
Solution:
We know that the sum of the first n odd natural numbers is n2 . Since,49 = 72
∴ 49 = sum of first 7 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13
(ii) Express 121 as the sum of 11 odd numbers. Solution:
Since, 121 = 112
∴ 121 = sum of first 11 odd natural numbers = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21
9. How many numbers lie between squares of the following numbers?
i. 12 and 13
ii. 25 and 26
iii. 99 and 100
Solution:
Between n2 and (n+1)2, there are 2n non–perfect square numbers.
i. 122 and 132, there are 2×12 = 24 natural numbers.
ii. 252 and 262, there are 2×25 = 50 natural numbers.
iii. 992 and 1002, there are 2×99 =198 natural numbers.
Exercise 6.1 of NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots is based on the following topics:
- Introduction to squares and square roots
- What are square numbers?
- Properties of Square Numbers
- Some interesting patterns
- Adding triangular numbers
- Numbers between square numbers
- Adding odd numbers
- A sum of consecutive natural numbers
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